Another way to do this: Choose two points on the line. For example, if t= 0 then (x, y, z) = (3, -2, -1) + (1, -2, 2)0= (3, -2, -1) and if t= 1, (x, y, z) = (3, -2, -1) + (1, -2, 2)1= (4, -4, 1). That gives (-3, 4, 2), (3, -2, -1), and (4, -4, 1) as three points in the plane and we can use the standard method of finding the equation of the plane that includes three given points.

Now, the vector from (-3, 4, 2) to (3, -2, -1) is 6i- 6j- 3k and the vector from (-3, 4, 2) to (4, -4, 1) is 7i- 8j- k. The cross product of those two vectors is a normal vector to the plane and its equation can be written A(x-x0)+ B(y-y0)+ C(z-z0)= 0 where Ai+ Bj+ Ck is a normal vector and (x0, y0, z0) is a point in the plane.